Question: The sum of two positive integers is 11, and the positive difference between their squares is 55. What is the positive difference of the two integers?
Solution: Let the two positive integers be $a$ and $b,$ with $a>b.$ So, \[a+b=11\] and \[a^2-b^2=55.\]$a^2-b^2$ can be factored as $(a+b)(a-b).$ Substituting $11$ for $(a+b)$ yields $11\cdot(a-b)=55$, implying that $a-b=55\div11=\boxed{5}.$